Note that the first part of the equation is the projection of the modulus of - B_B onto the vertical axis. And the second part is the projection of the modulus of B_C also on the vertical axis. On the horizontal axis the two components cancel because they have opposite signs. Therefore, calculating the expression above, remembering that cos 30° = 0.866, we arrive at the result of the resulting phasor, or

Therefore, we conclude that the resulting phasor is, in module, equal to 1.5 B_max. It should be noted that for ωt = 0° the resulting phasor makes an angle of 90° with the horizontal.

Now, let's calculate the resulting phasor for values ​​of ωt other than zero. Enjoy Figure 106-03 where we show the formation of the resulting phasor for ωt = 30°, ωt = 60° and ωt = 90°. Notice that regardless of the considered angle, the resulting phasor value, B_R, in module, is always equal to 1.5B_max. But, the reader should be aware of the fact that the resulting phasor angle changes to as we increase the viewing angle. And more: since we are using the sequence direct or positive, notice that the resulting phasor is rotating clockwise. We easily conclude that if we use the sequence inverse or negative, the resulting phasor rotates counterclockwise.

Some authors, in their books, want to show that the resulting phasor B_R rotates counterclockwise. Although they define the phases in direct sequence, when preparing the graph they use the inverse sequence, as this sequence is what allows the resulting phasor to be counterclockwise. On this site, we chose to show that with the direct sequence the resulting phasor rotates clockwise.

Keep in mind that to find the resulting phasor value for ωt = 30°, we use trigonometry, remembering that | - B_B | = B_max, B_C = 0.5 B_max and B_A = 0.5 B_max. So we can write

Taking into account that 0.5 B_max cos (60°) = 0.25 B_max, then the end result is

That is, once again we find that the module of the resulting phasor has a constant value equal to 1.5 B_max .

In Figure 106-04 we are showing the formation of the rotating phasor when ωt = 120°, ωt = 150° and ωt = 180°. From this moment on, we will present the graphs of the formation of the rotating phasor, without much explanation additional, as the graphics themselves are self-explanatory. For didactic reasons, and because this subject is not included in the most books dealing with machines, we will show all twelve graphs so that the reader understands how to process the formation of the rotating field. It is extremely important to understand this part so that the reader can understand how a machine can enter the rotation process.

In fact, we present 13 graphs, including the one for ωt = 360° so that the reader will notice that the first and tenth third graph are absolutely the same, showing a complete turn of the rotating phasor. For subsequent angles there will be a repeat of the above. Keep in mind that here are the thirteen main angles for a complete understanding of how the resulting phasor rotates 360°. It is clear that for a power system operating at a frequency 60 Hz, the rotating phasor will perform 60 full turns per second. If the frequency were 50 Hz then that would be 50 full turns per second. And so on.

Back to Top

As we stated earlier in item 3 (to see click here!), the synchronous motor does not have the capacity starting simply by applying the working voltages. Remember that the synchronous motor rotor is made up of several silicon-iron sheets juxtaposed to form a considerable volume. As a consequence, this rotor has a reasonable value mass. This creates a moment of inertia. And, as we know, to change the inertia of a body we must apply a force in an appropriate way. The greater the moment of inertia, the greater the value of the force must be.

Imagine a synchronous motor at rest. We connect a DC voltage in the field winding and a three-phase set of AC voltages in the stator winding producing a three-phase magnetic flow of currents in the windings. As we studied in the previous item, this procedure will create a rotating uniform magnetic induction field, which we will call B_S. Thus, there are two magnetic fields present in the machine and the rotor field will tend to line up with the stator field. As the stator's magnetic field is rotating, the rotor's magnetic field (and the rotor itself) will constantly try to line up. The greater the angle between the two magnetic fields (up to a certain maximum value), the greater the torque on the machine rotor. The motor rotor is stationary and, therefore, the magnetic field produced by the rotor, which we will call B_R, is stationary. The magnetic field B_S of the stator is rotating inside the motor at synchronous speed. However, the synchronous motor fails to start. Why?

To understand what is happening, let's analyze the equation of the torque induced in the machine by the two existing magnetic fields. The equation is shown below. See eq. 106-02.

Where the variables involved are:

Notice that in this equation, there is a cross product. The eq. 106-03 presents this same equation in its scalar form.

Where the variables involved are:

Let's start our analysis by assuming a network with a frequency of 60 Hz and looking at Figure 106-07. In this figure we show the rotor with its field winding and the phase shift between the magnetic induction field of the rotor, B_R, and the stator magnetic induction field, B_S, for different instants of time. We are assuming that the machine rotor is stopped.

At the instant t = 0 s, when we connect the voltages in the windings, we have the two field phasors aligned, that is, the angle between them is 0°. To calculate the induced torque we use the eq. 106-03. However, we know that sin 0° = 0. Soon, we conclude that the value of the induced torque, at that instant, is null.

At the instant t = 1/240 s, the stator field, B_S, is 90° ahead of the field of the rotor B_R. In this case, sin 90° = 1 and we have an induced torque. According to the right hand rule, torque is entering the page.

For the case t = 2/240 s, we see that the angle between the two phasors is 180°. And since sin 180° = 0, then the torque is null.

And at the instant t = 3/240 s, the angle between the phasors is 270° and as sin 270° = -1 , then we have a torque induced and according to the right hand rule the torque is leaving the page.

After this analysis, we realized that the times are very short and it is not possible for the rotor to follow the phasor speed of the field stator induction. And as a consequence of what was said above, we can say that in an electric cycle the average torque can be considered null. As a result, the engine vibrates, does not produce any turning, and furthermore, it overheats.

So the question is:

How to start a synchronous motor ?

To answer this question, there are three alternatives as described below.

We will study separately each of the alternatives.

Back to Top

If the magnetic field of the stator of a synchronous motor rotates at a low enough speed, there will be no problem for the rotor accelerates and synchronizes with the magnetic field. The speed of the stator magnetic field can then be gradually increased until reaching a speed close to the synchronism speed. During this process, you should not place load on the motor shaft. From that moment on, it is possible to turn off the variable frequency system and turn on the winding of the stator to the three-phase power system. It is now possible to add load to the motor shaft.

Currently, solid-state motor controllers can be used to convert a constant input frequency into any desired output frequency. With the development of modern variable frequency drive packages in solid state, it became perfectly possible to continuously control the electrical frequency applied to the motor, running all values ​​from a fraction of a hertz to above the full nominal frequency. if such a variable frequency drive unit is included in a drive circuit. motor control to have speed control, synchronous motor starting becomes very easy. Just adjust the frequency for a very low starting value and then raise it to the desired operating frequency for normal operation.

When a synchronous motor operates at a speed lower than the rated speed, its internal generated voltage E_A will be smaller than normal. if the value of E_A is reduced, then the terminal voltage applied to the motor must also be reduced to keep stator current at safe levels. The voltage at any variable frequency drive or starter circuit should vary approximately linearly with the applied frequency.

The frequency converter is a high cost unit and we must carefully analyze its use. However, if the synchronous motor have to operate at variable speeds, this method is recommended.

Back to Top

The second method of starting a synchronous motor is by coupling it to a motor external starting and driving the synchronous machine to rated speed with the external motor. Next, the synchronous machine can be placed in parallel with the power system as a generator and the starter motor can be decoupled from the machine shaft. Upon reaching rated speed, the match can be turned off. Once the parallel input is complete, then the motor synchronous can receive load in the normal way.

The starter only needs to overcome the inertia of the machine synchronous to empty. No load is applied until the motor is paralleled with the power system. As only the inertia of the motor needs to be overcome, the starter may have a much lower rated characteristic than that of the synchronous motor which he is starting.

In many synchronous motors, from medium to large, an external motor of starting or the use of the exciter may be the only possible solutions, because the power systems to which they are connected are not capable of handling the inrush currents necessary for damper windings to be used. Damper windings are our next topic.

Back to Top

The most popular way to start a synchronous motor is to use damper windings. These windings are special bars placed in slots cut into the rotor face of a synchronous motor and then short-circuited at each end by a large short circuit ring.

Assume initially that the main field winding of the rotor is off and that a three-phase set of voltages is applied to the stator winding of that machine. When power is initially applied at time t = 0 s, assume that the induction field B_S is vertical. When the magnetic induction field B_S rotates counterclockwise, it induces a voltage in the damper winding bars that is given by eq. 106-04.

Where the variables involved are:

Thus, when the rotor is turning, it induces a voltage between the ends of the bar. This voltage on the bars generates a current that will produce a magnetic induction field, which we will call B_W, which will interact with the stator magnetic induction field. In this way, a resultant induced torque will appear in the bars (and in the rotor) in the direction counter-clockwise. The value of the conjugate is given by eq. 106-02, already studied (above). Continuing with the analysis, we conclude that, sometimes the magnetic induction field is null, sometimes it is different from zero, but it always points in the same direction. That causes the average torque on the rotor to be different from zero. Therefore, the rotor is able to come out of rest and gain velocity. It is important to understand that the rotor cannot reach synchronous speed, because if the rotor is rotating at same speed as the stator magnetic induction field, the relative speed between the rotor and B_S is zero. Now, if the relative velocity is null, then by eq. 106-04 the voltage induced in the bars will be null, not generating the magnetic induction field B_W. Obviously, by eq. 106-02 the torque will be null and there will be no reason for the rotor to keep turning.

In a real machine, the field winding circuits cannot be open-circuited during the starting procedure, because in that case, very high voltages would be produced in them during the match. If the field windings are short-circuited during start-up, dangerous voltages will not be produced and, in reality, the field current induced in it will contribute with an additional torque to the starting of the motor.

It is then possible to start a synchronous motor with damper bars following this procedure:

Back to Top

One of the important features of damper windings is a significant improvement in the stability of the motor. The stator magnetic field rotates at a constant speed n_sinc, which changes only if the system frequency suffers some variation. If the rotor rotates at speed n_sync, the damper windings will not have any induced voltage. If the rotor turns slower than n_sync, then there will be relative motion between the rotor and the field stator magnetic field and a voltage will be induced in the windings. This voltage produces a current flow, which produces a field magnetic.

The interaction of the two magnetic fields produce a torque that tends to increase the speed of the machine again. On the other hand, if the rotor turns faster than the field of the stator, then a torque will be produced that will try to reduce the rotor speed. So, the conjugate produced by damper windings speeds up slow machines and slows down fast machines.

Therefore, these windings tend to dampen the load and other transients of the machine. For this reason, these windings are called windings shock absorbers. Such windings are also used in synchronous generators that operate in parallel with other infinite bus generators.In this case, the windings are used in a similar function of stabilization. If there is a torque variation on the generator shaft, its rotor will momentarily accelerate or will slow down and these changes will be opposed by the damper windings. These windings improve the overall stability of power systems by reducing power and torque transients.

The damper windings are responsible for the majority of the subtransient current of a synchronous machine in idle condition. electrical fault. a short circuit across the terminals of a generator is simply another form of transient and the damper windings react very quickly to it.

Back to Top

A synchronous motor is the same as a synchronous generator in all respects, except that the direction of power flow is inverted. How is this sense inverted, it can be expected that the direction of current flow in the stator will also be inverted. Therefore, the equivalent circuit of a synchronous motor is exactly the same as the equivalent circuit of a generator synchronous, except for the fact that the direction I_a reference to be inverted. In Figure 106-08 we can see the equivalent circuit of a motor synchronous.

Of course, a synchronous motor can be connected in delta or star configuration. As I_a changed its meaning, so the equations that govern the synchronous motor also change the sign. Then, we have:

That is, they are the same equations studied for a synchronous generator, differing only by the sign of the current term. And to determine the value of I_a, starting from eq. 106-05 we arrive at eq. 106-07, below.

The electrical model of a synchronous motor is the same as that of a synchronous generator, differing by the direction of current I_a, as this is opposite in the engine. To see the model in more detail click here!.

Back to Top

Based on Figure 106-08 we can draw a phasor diagram of the synchronous motor as shown in Figure 106-09.

Based on the phasor diagram shown in Figure 106-09 we can write that:

Note that eq.106-08 is an alternative form of eq.106-06. The diagram above shows that the power angle is negative and, consequently, the real or active power is also negative, meaning that the machine absorbs power from the power supply. It is easy to conclude that the smaller the power angle, δ, the smaller will be the transfer of real power.

Back to Top

Synchronous motors supply power to loads, which are basically devices that run at constant speed. Usually, engines are connected to power systems that are much larger than themselves, so the Power systems act as infinite buses for motors. This means that the terminal voltage and system frequency will be constant regardless of the amount of power demanded by the engine. The rotation speed of the motor is synchronized with the rate of rotation of the magnetic fields and, in turn, the rate of rotation of the applied magnetic fields is synchronized with the applied electrical frequency, so that the speed of the motor synchronous will be constant regardless of the load.

The torque versus speed characteristic curve is shown in Figure 106-10. The steady state speed of the engine is constant from no-load to the maximum torque that the motor can supply, so that the speed regulation of that motor is 0 %.

The torque in a synchronous motor is given by eq. 106-09. Note that the torque is directly proportional to the sine of the power angle, δ.

Note by eq. 106-09, that the maximum torque occurs when the power angle δ is equal a 90°, because sin 90° = 1. However, the normal conjugates to full load are much lower than this value. In fact, the maximum torque can be typically three times the full load torque of the machine.

When the shaft torque of a synchronous motor exceeds the maximum torque, the rotor may lose synchronism with stator and net magnetic fields. Instead, the rotor begins to slip, lagging behind. The loss of sync after the maximum torque is exceeded it is known as sliding poles.

From eq. 106-09 and based on the relationship of eq. 105-08 (see here!), we can write the equation that calculates the power converted into mechanical power on the axis of the synchronous motor. Look the eq. 106-10, below.

We know that the greater the field current I_F, the greater the induced voltage E_A and, consequently, the greater the torque developed by the motor. This is what tells us eq. 106-09. Thus, it is possible to conclude that there are advantages, from the point of view of stability, when working with high field currents I_F or, consequently, with high induced voltage E_A.

Back to Top

Let's assume a synchronous motor running at synchronous speed with a given load, initially with a leading power factor, as shown in Figure 106-11. In this case, the motor develops enough torque to keep the system rotating at synchronous speed.

Now, let's assume that there was an increase in the load coupled to the motor shaft. Initially, the motor will reduce its speed. By doing so, the δ torque angle becomes larger and the induced torque increases. the increment in the induced torque will accelerate the motor, which will turn again at synchronous speed, but with a larger conjugate angle.

The generated internal voltage E_A, given by eq. 102-02 (see here!), shows that the internal voltage only depends on the field current (since it has an influence on the Φ flow value) and the machine speed. The velocity is conditioned to be constant by the supply system and, as there were no changes in the field circuit, the field current is also constant. Therefore, |E_A| must be constant when the load changes. In the graph shown in Figure 106-12, the lengths proportional to the power (E_A sin δ and I_a cos θ) will increase, but the modulus of E_A should remain constant. When the load increases, E_A moves down as shown in the graph. When E_A moves further and further down, the term j X_S I_a must increase so that it can go from the end of E_A up to V_T and consequently the armature current I_a will also increase. Watch that the angle θ (angle between I_a and V_T) of power factor also changes, becoming increasingly less early and then more and more late.

From everything that has been presented here, it can be concluded about the effects of the increase in load, under conditions of excitation constant (disregarding the effects of the armature reaction), that:

Back to Top

In a synchronous motor there is another magnitude that can be adjusted easily, namely its field current. Let's imagine a synchronous motor initially operating with a lagging power factor, situation 1 as shown in Figure 106-13. Now let's increase the field current and see what happens. Note that an increase in the field current generates a rise in the module of E_A, but does not affect the active power delivered to the engine. The power delivered to the motor changes only when the load varies. Since a change in I_F does not affect the speed of the motor shaft and, as the load coupled to the shaft does not change, the supplied active power does not change.

Of course, V_T is also constant, because it's kept that way by the power source that powers the motor. The power proportional lengths in the diagram phasorial (E_A sin δ and I_a cos θ) must therefore be constants. When the current of field is increased, the voltage E_A must increase, but it can only do this by following the constant power line. This effect is shown in Figure 106-13.

When the internal voltage phasor projection E_A onto V_T (E_A cos δ) is smaller than V_T itself, a synchronous motor has an armature current delayed and consumes reactive power Q. In this case, the field current I_F is small and we say that the motor is underexcited. This means that the synchronous motor is a combination of inductor and resistor (case of E_A1 and I_a1 in the graph above).

When the internal voltage phasor projection E_A onto V_T (E_A cos δ) is greater than V_T itself, a synchronous motor has an armature current lead and consumes negative reactive power -Q or, what is the same thing, supplies reactive power Q to the system. In this case, the field current I_F is large and we say that the motor is over-excited. This means that the synchronous motor is a combination of capacitor and resistor (case of E_A3 and I_a3 in the graph above).

When the internal voltage phasor projection E_A onto V_T (E_A cos δ) is equal to V_T, a synchronous motor has an armature current at phase with V_T, and consumes active power P. In this case, the field current I_F is at an intermediate value between the two previous cases. This means that the synchronous motor is operating as a resistor (case of E_A2 and I_a2 in the graph above).

A plot of I_a versus I_F for a synchronous motor is shown in Figure 106-14 . This graph is called the V curve of a synchronous motor, for the obvious reason that its shape is like the letter V.

There are several V-curves drawn, corresponding to different levels of active power. For each curve, the minimum armature current I_a occurs with unity power factor, when only active power is being supplied to the engine. At any other point on the curve, some reactive power is also being supplied to or from the motor. For smaller field currents than the value corresponding to I_a minimum, the armature current is lagging, consuming reactive power Q. For field currents greater than the value corresponding to I_a minimum, the armature current is leading, providing reactive power Q to the power system, like a capacitor would. Therefore, controlling the field current of a synchronous motor, we will be able to control the supplied reactive power or consumed by the power system.

Back to Top

A synchronous motor, purchased to drive a load, can be operated over-excited in order to supply power. reactive Q for a power system. In In fact, in the past, a synchronous motor was bought to run without load, simply to perform power factor correction. The phasor diagram of a synchronous motor, running over-excited at no-load, is shown in Figure 106-15.

Since there is no power being drawn from the motor, the proportional phasor lengths to the power (E_A sin δ and I_a cos θ) are null. We know that from Kirchhoff's voltage law for a motor synchronous, considering that X_S >> R_a, is given by:

The term j X_S I_a will point to the left and therefore the armature current I_a will point up. If V_T and I_a are examined, the voltage and current relationship between them it will be like that of a capacitor. From a power system point of view, an engine synchronous to empty overexcited looks exactly like a big capacitor.

Some synchronous motors used to be sold specifically for power factor correction. These machines had axles that didn't even come out. of the engine frame and no load could be attached to them, even if that was desired. These special purpose synchronous motors were often called synchronous capacitors.

Currently, conventional static capacitors are more cost-effective to buy and use than synchronous capacitors. Nonetheless, some synchronous capacitors may still be in use in older industrial installations.